Alexander — Maximum Bending -moments on Short Girders. 21 



distance froin 2? to fr is {Z- 6), and the bending-moment at Fi is {Z- 6) (42 -Z) 

 - 9 X 7. Equating these, gives Z=3o feet, which is numerically equal to the 

 sum of the loads on the first four wheels. In this way it is shown that the 

 first parabola is above all the others for the first 5 feet of the span, the second 

 parabola for the next 5 feet, the third parabola for the next 1 1 feet : that 

 is, each wheel commands a portion of the span or a " field," which is the same 

 fraction of the span that the load on that wheel is of the total load. As the 

 load and span have been taken numerically equal, the five fields into which 

 the span is divided are 5, 5, 11, 12, and 9 feet respectively. At any point 

 in any field the maximum bending-moment occurs when the commanding 

 weight is over that point. In the two fields, one on each side of the middle 

 of the span, there is a maximum of the maxima S^Az, 2 feet to the left of 

 the middle of the span, and SiA^ 3 feet to its right. Or in symbols 



zMi = 256 foot-tons and iM.^ = 261 foot-tons. 



As the second of these is the greater, we have called the wheel ^4 = 12 tons 

 the ruling-wheel of the locomotive when riding on a girder or bridge of span 

 42 feet. 



To find a uniformly distributed load which will give the same bending- 

 moment, 261 foot- tons, at the same point, 3 feet to the right of the middle of 

 the span, we can assume a parabolic locus like BAoC standing on the span, 

 but passing through Ai. This locus representing the bending-moments due 

 to a uniformly distributed load of intensity w tons per foot, gives if we put 

 C3 21 the half span, 



S,A, = ~xBF,x OF, = I (c + 3)(c - 3) ; 



and equating this to 261, we get vj = 1-209 tons per foot of span. 



The author's original graphical solution (fig. 2) was drawn with such a 

 template sliding on a tee-square, and pushed with its vertex up the axes of 

 the five parabolic loci, one after another, taking care to make these loci 

 cross each other on the verticals through the junctions of the fields. 



On fig. 3 we now show a method of drawing the five loci without considering 

 the junctions of the fields, It does not give so good a definition, but is 

 instructive. Thus, the locomotive is to be fixed with its centre of gravity 

 over the middle point of the span, and the instantaneous bending-moment 

 polygon BabcdeC drawn to the vertical scale, upon which the height of A^^ 

 reads 441 foot-tons. Now, on page 19 it was pointed out that the locus 

 whose vertex is Ai has equal heights at 6 feet to the right of the middle point 

 of the span and at the middle point itself. But on fig. 3 the wheal Wi is 

 6 feet to the right of the middle of the span. So that the perpendicular 



